Name: Light Refraction Read from Lesson 1 of the Refraction and Lenses chapter at The Physics Classroom: http://www.physicsclassroom.com/class/refrn/u14l1a.html http://www.physicsclassroom.com/class/refrn/u14l1b.html http://www.physicsclassroom.com/class/refrn/u14l1c.html http://www.physicsclassroom.com/class/refrn/u14l1f.html MOP Connection: Refraction and Lenses: sublevels 1 and 2 1. Write a one-word synonym for refraction. 2. Refraction occurs when light crosses the boundary between one material and another material. What is the primary cause for this refracting of light upon crossing a boundary? The diagram below shows the path of a light ray as it travels through air, across the air-water boundary, and through the water. Use the diagram to answer questions #3-#6. 3. On the diagram, label... the air-water boundary with a B the normal line with an N the incident ray with an I the refracted ray with an R the angle of incidence with a θi the angle of refraction with a θr 4. How many media are there in this diagram? Name them. 5. What is meant by the term "medium" in this context? 6. Place a noticeable dot at the location where refraction of light takes place. 7. For the three situations below, draw a normal line and measure and record the angles of incidence and the angles of refraction. θi = θi = θi = θr = θr = θr = 8. As light passes from one medium into another, it refracts. There is only one condition in which light will cross a boundary but not refract. State this condition. The Physics Classroom, 2009 Page 1
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Name: Direction of Bending Read from Lesson 1 of the Refraction and Lenses chapter at The Physics Classroom: http://www.physicsclassroom.com/class/refrn/u14l1d.html http://www.physicsclassroom.com/class/refrn/u14l1e.html http://www.physicsclassroom.com/class/refrn/u14l1f.html MOP Connection: Refraction and Lenses: sublevels 2 and 3 1. The optical density is the property of a medium which provides a relative measure of the speed at which light travels in that medium. Light travels (fastest, slowest) in media with a greater optical density. 2. Every transparent material is characterized by a unique index of refraction value (n). The index of refraction value is a numerical value which provides a relative measure of the speed of light in that particular material. Light travels (fastest, slowest) in media with a higher index of refraction value. 3. The speed of light (v) in a material is determined using the speed of light in a vacuum (c) and the index of refraction (n) of the material. Calculate the speed of light in the following materials. v = a. water (n = 1.33): b. glass (n = 1.50): c. ice (n = 1.31): d. diamond (n = 2.42): c n = 3.00 x 10 8 m/s n 4. When light passes into a medium in which it travels faster, the light will refract the normal. When light passes into a medium in which it travels slower, light will refract the normal. a. towards, away from b. away from, towards 5. When light passes into a medium which is more optically dense, the light will refract the normal. When light passes into a medium which is less optically dense, the light will refract the normal. a. towards, away from b. away from, towards 6. Consider the refraction of light in the five diagrams below. In which case is the light bending towards the normal line? Circle all that apply. Consider the diagram at the right in answering the next four questions. 7. There are (1, 2, 3,...) media shown in the diagram. 8. There are (1, 2, 3,...) boundaries shown in the diagram. 9. Light must travel in medium 1 compared to medium 2. a. slower b. faster c. insufficient info 10. Light must travel in medium 2 compared to medium 3. a. slower b. faster c. insufficient info The Physics Classroom, 2009 Page 3
11. In each diagram, draw the "missing" ray (either incident or refracted) in order to appropriately show that the direction of bending is towards or away from the normal. 12. A ray of light is shown passing through three consecutive layered materials. Observe the direction of bending at each boundary and rank the three materials (A, B and C) in order of increasing index of refraction. smallest < < largest 13. Arthur Podd's method of fishing involves spearing the fish while standing on the shore. The apparent location of a fish is shown in the diagram below. Because of the refraction of light, the observed location of the fish is different than its actual location. If Arthur is to successfully spear the fish, must he aim at, below, or above where the fish appears to be? Explain. The Physics Classroom, 2009 Page 4
Name: Snell's Law Read from Lesson 2 of the Refraction and Lenses chapter at The Physics Classroom: http://www.physicsclassroom.com/class/refrn/u14l2b.html http://www.physicsclassroom.com/class/refrn/u14l2c.html http://www.physicsclassroom.com/class/refrn/u14l2d.html MOP Connection: Refraction and Lenses: sublevel 4 Math Review: To find an angle measure for which the sine value is 0.8660, enter 2nd Sin (0.8660) into your TI graphing calculator and press the Enter key. The angle is 59.99... degrees. 1. Use Snell's law to solve the following physics word problems. PSYW a. An incident ray in air (n=1.0) is approaching the boundary with an unknown material at an angle of incidence of 61.6. The angle of refraction is 41.4. Determine the index of refraction of the unknown material. b. An incident ray in air (n=1.0) is approaching the boundary with glass (n = 1.52) at an angle of incidence of 32.5. Calculate the angle of refraction. Draw the refracted ray on the diagram at the right. Air Glass 2. For the following two situations, measure and record θi, calculate θr, and draw in the refracted ray with the calculated angle of refraction. PSYW The Physics Classroom, 2009 Page 5
3. The diagram at the right shows a light ray entering a rectangular block of unknown material and subsequently exiting the block on the opposite side. The path of the light ray through the block is shown. Determine the index of refraction of the unknown material. Perform two calculations - one for each boundary - using Snell's law and the measured angles. PSYW 4. Cal Culator is performing experiments to determine the index of refraction of two unknown materials. Cal determines that the light follows the paths as shown on the diagrams below. Use this path, a protractor, a calculator and Snell's Law to determine the index of refraction of the unknown material. Show all your work in the space beside the diagram. The Physics Classroom, 2009 Page 6
Name: Total Internal Reflection Read from Lesson 3 of the Refraction and Lenses chapter at The Physics Classroom: http://www.physicsclassroom.com/class/refrn/u14l3a.html http://www.physicsclassroom.com/class/refrn/u14l3bhtml http://www.physicsclassroom.com/class/refrn/u14l3c.html MOP Connection: Refraction and Lenses: sublevels 5 and 6 Background: Whenever a light ray reaches the boundary with a transparent medium, a portion of the light energy is transmitted across the boundary and appears as a refracted ray; and a portion of the energy remains within the original medium and appears as a reflected ray. The path of the refracted ray follows Snell's law. The path of the reflected ray follows the law of reflection. The amount of light energy which is reflected and transmitted is dependent upon the angle of incidence. At certain angles, all of the light is reflected (none is transmitted) and remains inside the original medium. This is known as total internal reflection (TIR). 1. Light will undergo total internal reflection only when it is. Choose two. a. in the less dense medium traveling towards the more dense medium b. in the more dense medium traveling towards the less dense medium c. in the medium where it travels slowest, moving towards the medium where it travels fastest d. in the medium where it travels fastest, moving towards the medium where it travels slowest 2. Total internal reflection is most likely to occur when. a. the angles of incidence are smaller (e.g., close to 0 degrees) b. the angles of incidence are greatest (e.g., close to 90 degrees) Complete the following blanks by answering questions #3-#4: The critical angle is the angle of (#3) which causes light to (#4). 3. Referring to the statement above: a. incidence b. refraction c. reflection 4. Referring to the statement above: a. cross the boundary without refracting b. undergo refraction at the same angle as the angle of incidence c. refract at an angle of refraction of 90 degrees d. reflect at the same angle as the angle of incidence The next three questions focus on the brightness of the reflected and refracted rays and the dependency of the brightness upon the angle of incidence. 5. Consider the diagram at the right for rays A, B, C, and D incident upon a water-air boundary. The corresponding refracted rays are shown. Draw the corresponding reflected rays and label them as A'', B'', C'', and D''. 6. As the angle of incidence is gradually increased, more and more of the energy from the incident ray goes into the reflected ray, while less and less of the energy goes into the refracted ray. Based on this fact, which one of the refracted rays in the diagram would be brightest and which one would be dimmest? Brightest: Dimmest: 7. For incident ray C, the angle of refraction is 90. The refracted ray C has the smallest amount of energy of any refracted ray. Thus, it would be an extremely "dim" light ray. What is the angle of incidence for ray C called? The Physics Classroom, 2009 Page 7
8. The critical angle for an air (n=1.0) - lucite (n=1.4) boundary is approximately 46 degrees. Which of the following diagrams depict incident rays which would undergo total internal reflection (TIR) at the angle shown? Circle all that apply. 9. Calculate the critical angle for the... a.... air (n = 1.00) - water (n = 1.33) boundary: b.... air (n = 1.00) - diamond (n = 2.42) boundary: c.... water (n = 1.33) - glass (n = 1.50) boundary: 10. a. Calculate the critical angle for the boundary between glass (n = 1.50) and diamond (n = 2.42). PSYW b. On the diagram at the right, draw an incident ray which approaches the boundary with an angle equal to the critical angle. Label the incident ray as A. Draw the corresponding refracted ray and label the ray as B. c. Draw an incident ray which would approach the boundary at an angle greater than the critical angle. Label this incident ray as C. Glass (n=1.50) Diamond (n=2.42) 11. Diamonds are usually cut with a shape similar to that shown at the right. Kent Affordit is preparing to propose to Amanda Befrendswyth. In an effort to save money, Kent asked the jeweler to remove the bottom portion of the gem. Kent reasoned that since it was not visible, its removal would have little consequence to its ultimate appearance. Explain why Kent never did get engaged to Amanda. Finally, draw the path of the given incident ray in each diamond. The Physics Classroom, 2009 Page 8
Name: Lenses Read from Lesson 5 of the Refraction and Lenses chapter at The Physics Classroom: http://www.physicsclassroom.com/class/refrn/u14l5a.html http://www.physicsclassroom.com/class/refrn/u14l5b.html http://www.physicsclassroom.com/class/refrn/u14l5c.html MOP Connection: Refraction and Lenses: sublevel 7 1. Converging lenses are at the center and at the edges. a. thickest, thinnest b. thinnest, thickest 2. Diverging lenses are at the center and at the edges. a. thickest, thinnest b. thinnest, thickest Consider the diagram at the right in answering the next two questions. 3. List the letters of all the converging lenses. 4. List the letters of all the diverging lenses. ' 5. Use refraction principles to sketch an approximate path of light as it enters and exits the lens. Think FST and SFA. Trace the path of the rays into, through and out of the lens. Repeat the procedure for the light rays exiting the lens and trace the emerging light rays. Place arrowheads on all light rays. 6. Explain why lenses (like the one on the left above) are called "converging" lenses. 7. Converging lenses will have (positive, negative) focal lengths. Diverging lenses will have (positive, negative) focal lengths. The Physics Classroom, 2009 Page 9
8. The diagram below shows an arrow object positioned in front of a converging and a diverging lens. Three incident rays are shown. Construct the corresponding refracted rays. Show arrowheads. 9. State the three rules of refraction for converging lenses: #1: #2: #3: 10. State the three rules of refraction for diverging lenses: #1: #2: #3: 11. The diagrams below depict the refraction of light through various lenses. List the diagrams that show the proper refraction of light. For those which show the improper refraction of light, either correct the diagrams by showing the proper refracted rays or explain what is wrong with the refracted rays. The Physics Classroom, 2009 Page 10
Name: Ray Diagrams for Converging Lenses Read from Lesson 5 of the Refraction and Lenses chapter at The Physics Classroom: http://www.physicsclassroom.com/class/refrn/u14l5da.html http://www.physicsclassroom.com/class/refrn/u14l5db.html MOP Connection: Refraction and Lenses: sublevels 8 and 9 For the following lenses and corresponding object positions, construct ray diagrams. Then describe the Location of the image, Orientation (upright or inverted) of the image, the relative Size of the image (larger or smaller than object), and the Type of image (real or virtual). For Case 4, merely construct the ray diagram. NOTE: 1) All light rays have arrowheads which indicate the direction of travel of the ray. 2) Always draw in the image once located (an arrow is a good representation). 3) Exactness counts. Use a straight-edge and be accurate. Case 1: If the object is located beyond 2F: Description of Image: Location: O: Upright or Inverted S: Magnified or Reduced T: Real or Virtual Case 2: If the object is located at 2F: Description of Image: Location: O: Upright or Inverted S: Magnified or Reduced T: Real or Virtual The Physics Classroom, 2009 Page 11
Case 3: If the object is located between 2F and F: Description of Image: Location: O: Upright or Inverted S: Magnified or Reduced T: Real or Virtual Case 4: If the object is located at F: No Description Required Case 5: If the object is located between F and the lens: Description of Image: Location: O: Upright or Inverted S: Magnified or Reduced T: Real or Virtual The Physics Classroom, 2009 Page 12
Name: Ray Diagrams for Diverging Lenses Read from Lesson 5 of the Refraction and Lenses chapter at The Physics Classroom: http://www.physicsclassroom.com/class/refrn/u14l5ea.html http://www.physicsclassroom.com/class/refrn/u14l5eb.html MOP Connection: Refraction and Lenses: sublevels 10 and 11 For the following lenses and corresponding object positions, construct ray diagrams. Then describe the Location of the image, Orientation (upright or inverted) of the image, the relative Size of the image (larger or smaller than object), and the Type of image (real or virtual). NOTE: 1) All light rays have arrowheads which indicate the direction of travel of the ray. 2) Always draw in the image once located (an arrow is a good representation). 3) Exactness counts. Use a straight-edge and be accurate. Case 1: If the object is located far away from the lens: Description of Image: Location: O: Upright or Inverted S: Magnified or Reduced T: Real or Virtual Case 2: If the object is located nearby the lens: Description of Image: Location: O: Upright or Inverted S: Magnified or Reduced T: Real or Virtual The Physics Classroom, 2009 Page 13
Lenses and Mirrors - Applying Concepts 1. Light emanates in a variety of directions from the following point objects; some of this light is incident towards the mirror or lens. The behavior of a few such incident rays are shown below. Show how the third, fourth and/or fifth incident rays refract or reflect. Converging Lens Converging Lens Concave Mirror Diverging Lens 2. Several statements about images are given below. Identify which optical device applies to the given statement. Place the appropriate marks in the blanks. Mark all that apply. A = plane mirrors B = concave mirrors C = convex mirrors D = converging lenses E = diverging lenses a. Are capable of producing real images. b. Only produce virtual images. c. Are capable of producing enlarged images. d. Can only produce images which are smaller than the object. e. Capable of producing images the same size as the object. 3. Identify the following statements as being either true (T) or false (F). a. If reflected or refracted rays diverge, there is no image. b. If an object is located in front of a focal point, there is no image. c. Virtual images cannot be seen. d. All images are formed by the actual convergence of reflected or refracted light. e. Just three rays of light from an object can intersect at the image location. The Physics Classroom, 2009 Page 14
Name: Lens Practice Read from Lesson 5 of the Refraction and Lenses chapter at The Physics Classroom: http://www.physicsclassroom.com/class/refrn/u14l5f.html Use the lens equation and magnification equation to solve the following problems. 1. Determine the image distance and image height for a 4.0-cm tall object placed 54.0-cm from a converging lens having a focal length of 18.0 cm. 2. Determine the image distance and image height for a 4.0-cm tall object placed 36.0-cm from a converging lens having a focal length of 18.0 cm. 3. Determine the image distance and image height for a 4.0-cm tall object placed 24.0-cm from a converging lens having a focal length of 18.0 cm. 4. Determine the image distance and image height for a 4.0-cm tall object placed 12.0-cm from a converging having a focal length of 18.0 cm. 5. A magnified, inverted image is located a distance of 32.0 cm from a converging lens with a focal length of 12.0 cm. Determine the object distance and tell whether the image is real or virtual. The Physics Classroom, 2009 Page 15
6. ZINGER: An inverted image is magnified by 2 when the object is placed 22 cm in front of a converging lens. Determine the image distance and the focal length of the lens. 7. A diverging lens has a focal length of -12.8 cm. An object is placed 34.5 cm from the lens's surface. Determine the image distance. 8. Determine the focal length of a diverging lens which produces an image which is 12.9 cm behind the lens when the object is 32.4 cm from the lens. 9. A 2.85-cm diameter coin is placed a distance of 31.4 cm from a diverging lens which has a focal length of -11.6 cm. Determine the image distance and the diameter of the image. 10. The focal point is located 20.0 cm from a diverging lens. An object is placed 12.0 cm from the lens. Determine the image distance. The Physics Classroom, 2009 Page 16